Question

The Ka of a monoprotic weak acid is 7.81 × 10-3. What is the percent ionization of a 0.130 M solution of this acid?

Answer 1

HA ---------------> H+ + A-

0.130                  0        0 ----------> initial

0.130-x               x         x ------------> equilibrium

Ka = x^2 / 0.130-x

7.81 x 10^-3 = x^2 / 0.130-x

x^2 + 7.81 x 10^-3 x - 1.02 x 10^-3 = 0

x = 0.0283

percent ionization =( x / C ) x 100

                              = (0.0283/ 0.130) x 100

                              = 21.8 %

percent ionization = 21.8 %